Bayes’ Theorem – IIT‑JEE (Main + Advanced) Study Notes
(Mathematics – Probability & Statistics)
Why it matters:
In JEE‑Main & JEE‑Advanced, problems on conditional probability and random experiments often require updating probabilities when new information is known. Bayes’ Theorem gives the precise way to do this.
1. Simple Explanation
| Concept | Meaning |
|---|---|
| Conditional Probability | Probability of an event A happening given that another event B has already happened. Denoted (P(A |
| Joint Probability | Probability that two events happen together: (P(A \cap B)). |
| Marginal Probability | Probability of an event irrespective of any other event: (P(A)). |
| Bayes’ Theorem | A formula that reverses a conditional probability: it gives (P(A |
Bayes’ Theorem (single event):
[ \boxed{P(A|B)=\dfrac{P(B|A),P(A)}{P(B)}} ]
- (P(A)) – Prior probability of A (before seeing B).
- (P(B|A)) – Likelihood – probability of observing B if A is true.
- (P(B)) – Evidence – overall probability of observing B (can be found using the Law of Total Probability).
- (P(A|B)) – Posterior probability of A after observing B.
2. Key Points & Formulae
| # | Item | Formula | Notes |
|---|---|---|---|
| 1 | Conditional Probability | (P(A | B)=\frac{P(A\cap B)}{P(B)}) |
| 2 | Law of Total Probability | If ({B_i}) partition the sample space: (P(A)=\sum_{i} P(A | B_i),P(B_i)). |
| 3 | Bayes’ Theorem (multi-partition) | (P(B_k | A)=\frac{P(A |
| 4 | Independence | (P(A\cap B)=P(A)P(B)) ⇒ (P(A | B)=P(A)). |
| 5 | Complement | (P(\overline{A})=1-P(A)). | Often used to compute evidences. |
| 6 | Bayes’ Theorem (with events of different types) | Same formula; just be careful with discrete vs continuous (density functions). | For JEE, always discrete problems. |
| 7 | Posterior Odds | (\frac{P(A | B)}{P(\overline{A} |
Quick Reference Sheet
| Symbol | Meaning | Example |
|---|---|---|
| (P(A)) | Prior probability of A | Probability a randomly chosen student is a boy |
| (P(B | A)) | Likelihood |
| (P(B)) | Evidence | Probability of drawing a red ball from the overall urn |
| (P(A | B)) | Posterior |
3. Worked Example (JEE‑Style)
Problem Statement
A box contains 3 red balls and 7 blue balls.
A second box contains 5 red balls and 5 blue balls.
A box is chosen at random; then a ball is drawn from that box.
- Let R = “the drawn ball is red.”
- Let B₁ = “Box 1 was chosen.”
- Let B₂ = “Box 2 was chosen.”
Question:
(a) Find (P(R)).
(b) If the drawn ball is red, find the probability that Box 1 was chosen, i.e., (P(B_1|R)).
Step‑by‑Step Solution
(a) Compute (P(R))
Use the Law of Total Probability:
[ P(R)=P(R|B_1)P(B_1)+P(R|B_2)P(B_2) ]
- (P(B_1)=P(B_2)=\frac12) (equal chance of selecting a box).
- (P(R|B_1)=\frac{3}{10}) (3 red out of 10).
- (P(R|B_2)=\frac{5}{10}=\frac12).
So,
[ P(R)=\frac12\cdot\frac{3}{10}+\frac12\cdot\frac12=\frac{3}{20}+\frac{1}{4}=\frac{3+5}{20}=\frac{8}{20}=\boxed{\frac{2}{5}} ]
(b) Find (P(B_1|R)) using Bayes’ Theorem
[ P(B_1|R)=\frac{P(R|B_1)P(B_1)}{P(R)} ]
Substitute the known values:
[ P(B_1|R)=\frac{\frac{3}{10}\cdot